If $x + \frac{1}{x} = 12$ find the value of $x - \frac{1}{x}$.

Given:

$x + \frac{1}{x} = 12$

To do:

We have to find the value of $x - \frac{1}{x}$.

Solution:

The given expression is $x + \frac{1}{x} =$. Here, we have to find the value of $x^4 + \frac{1}{x^4}$. So, by squaring the given expression and using the identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$, we can find the required value.

$(a+b)^2=a^2+2ab+b^2$.............(I)

$(a-b)^2=a^2-2ab+b^2$.............(II)

Let us consider,

$x + \frac{1}{x} = 12$

Squaring on both sides, we get,

$(x + \frac{1}{x})^2 = (12)^2$

$x^2+2\times x \times \frac{1}{x}+\frac{1}{x^2}=144$            [Using (I)]

$x^2+2+\frac{1}{x^2}=144$

$x^2+\frac{1}{x^2}=144-2$                       (Transposing $2$ to RHS)

$x^2+\frac{1}{x^2}=142$

Now,

$x^2+\frac{1}{x^2}=142$

Subtracting 2 from both sides, we get,

$x^2+\frac{1}{x^2}-2=142-2$

$x^2-2\times x \times \frac{1}{x}+\frac{1}{x^2}=140$                [Since $2\times x \times \frac{1}{x}=2$]

$(x-\frac{1}{x})^2=140$               [Using (II)]

Taking square root on both sides, we get,

$x-\frac{1}{x}=\sqrt{140}$

Hence, the value of $x-\frac{1}{x}$ is $\sqrt{140}$.